Develop Reference

How to write a function to evaluate this expression? [duplicate]

A friend of mine challenged me to write a function that works with both of these scenarios
add(2,4) // 6
add(2)(4) // 6
My instinct was the write an add() function that returns itself but I'm not sure I'm heading in the right direction. This failed.
function add(num1, num2){
if (num1 && num2){
return num1 + num2;
} else {
return this;
}
}
alert(add(1)(2));
So I started reading up on functions that return other functions or return themselves.
http://davidwalsh.name/javascript-functions
JavaScript: self-calling function returns a closure. What is it for?
JavaScript: self-calling function returns a closure. What is it for?
I am going to keep trying, but if someone out there has a slick solution, I'd love to see it!

I wrote a curried function whose valueOf() method and function context (this) are bound with the sum no matter how many arguments are passed each time.
/* add function */
let add = function add(...args) {
const sum = args.reduce((acc, val) => acc + val, this);
const chain = add.bind(sum);
chain.valueOf = () => sum;
return chain;
}.bind(0);
/* tests */
console.log('add(1, 2) = ' + add(1, 2));
console.log('add(1)(2) = ' + add(1)(2));
/* even cooler stuff */
console.log('add(1, 2)(3) = ' + add(1, 2)(3));
console.log('add(1, 2, 3)(4, 5)(6) = ' + add(1, 2, 3)(4, 5)(6));
/* retains expected state */
let add7 = add(7);
console.log('let add7 = add(7)');
console.log('add7(3) = ' + add7(3));
console.log('add7(8) = ' + add7(8));
The reason why both mechanisms are required is because the body of add() must use the called function's bound context in order to access the sum of the intermediate partial application, and the call site must use the valueOf() member (either implicitly or explicitly) in order to access the final sum.

There is an article on Dr.Dobs Journal about "Currying and Partial Functions in JavaScript" which describes exactly this problem.
One solution found in this article is:
// a curried add
// accepts partial list of arguments
function add(x, y) {
if (typeof y === "undefined") { // partial
return function (y) {
return x + y;
};
}
// full application
return x + y;
}

function add(num1, num2){
if (num1 && num2) {
return num1 + num2;
} else if (num1) {
return function(num2){return num1 + num2;};
}
return 0;
}

The concept that you're looking for is called currying and it has to do with function transformation and partial function application. This is useful for when you find yourself calling the same function over and over with mostly the same arguments.
An example of implementing add(2)(6) via currying would look something like this...
function add(x,y) {
if (typeof y === 'undefined') {
return function(y) {
return x + y;
}
}
}
add(2)(4); // => 6
Additionally, you could do something like this...
var add6 = add(6);
typeof add6; // => 'function'
add6(4); // => 10

var add = function(){
// the function was called with 2 arguments
if(arguments.length > 1)
arguments.callee.first_argument = arguments[0];
// if the first argument was initialized
if(arguments.callee.first_argument){
var result = arguments.callee.first_argument + arguments[arguments.length - 1];
arguments.callee.first_argument = 0;
return result;
}else{// if the function was called with one argument only then we need to memorize it and return the same function handler
arguments.callee.first_argument = arguments.callee.first_argument || arguments[0];
return arguments.callee;
}
}
console.log(add(2)(4));
console.log(add(2, 4));
An extended solution which depends on the environment:
function add(){
add.toString = function(){
var answer = 0;
for(i = 0; i < add.params.length; i++)
answer += add.params[i];
return answer;
};
add.params = add.params || [];
for(var i = 0; i < arguments.length; i++)
add.params.push(arguments[i])
return add;
}
console.log(add(2)(4)(6)(8))
console.log(add(2, 4, 6, 8));

We can use the concept of closures which is provided by Javascript.
Code snippet:
function add(a,b){
if(b !== undefined){
console.log(a + b);
return;
}
return function(b){
console.log(a + b);
}
}
add(2,3);
add(2)(3);

In general you need to have an agreement whether the function should return a function (for calling with more arguments) or the end result. Imagine the add function would have to work like this as well:
add(1, 2, 3)(4, 5) // -> 15
...then it becomes ambiguous, because you might want to call again:
add(1, 2, 3)(4, 5)(6) // -> 21
...and so add(1, 2, 3)(4, 5) should have returned a function, and not 15.
You could for instance agree that you have to call the function again, but without arguments, in order to get the numeric result:
function add(...args) {
if (args.length === 0) return 0;
let sum = args.reduce((a, b) => a+b, 0);
return (...args) => args.length ? add(sum, ...args) : sum;
}
console.log(add()); // 0
console.log(add(1,2,3)()); // 6
console.log(add(1,2,3)(4,5)()); // 15
console.log(add(1,2,3)(4,5)(6)()); // 21

One may think that he/she has to invoke the same function two times, but if you think deeply you will realize that the problem is pretty straight forward, you have to invoke the add function one time then you need to invoke what ever the add function returns.
function add(a){
return function(b){
return a+b;
}
}
console.log(add(20)(20));
//output: 40
you can return function as many as time you want. suppose for y = mx+c
const y= function (m){
return function(x){
return function (c){
return m*x+c
}
}
}
console.log(y(10)(5)(10));
//out put: 60

Random Fibonacci Generator

I'm trying to create a simple program in javascript where the Fibonacci square can be created by a random number sequence but I can't seem to connect both parts of my code. The first side being: the call for a random number and the second part: calculating the Fibonacci square.
var n = function getRandomNum() {
return Math.floor(Math.random()*100) +1;
}
function fib(x) {
if (x < 2) {
return x;
} else {
return fib(x - 1) + fib(x - 2);
}
}
console.log(fib(n));
Tell me where I'm going wrong. These are the errors I get when I run it.
RangeError: Maximum call stack size exceeded
at fib:7:13
at fib:11:12
at fib:11:12
at fib:11:12
at fib:11:12
at fib:11:12

Aside from not invoking the random number generator, you're using a very poorly optimized algorithm. If you think through all the redundant calls that need to take place, you'll see why the stack limit is reached.
var n = function getRandomNum() {
return Math.floor(Math.random() * 100) + 1;
}(); // <-- quick inline invocation... not normally how you'd use this.
console.log(n);
function fib(x) {
function _fib(x, a, b) {
if (x < 2) {
return a;
}
return _fib(x - 1, b, a + b);
}
return _fib(x, 0, 1);
}
console.log(fib(n));

Since you don't call n function, you should call it like the following.
var n = function getRandomNum() {
return Math.floor(Math.random()*100) +1;
}
function fib(x) {
if (x < 2) {
return x;
} else {
return fib(x - 1) + fib(x - 2);
}
}
console.log(fib(n));
But, there's a huge problem in your code, as #rock star mentioned, there's no any optimizing process in your code. That is why your code has caused the problem on memory leak
To avoid this, you can simply use memoization, click this link you don't have any clue on it.
Javascript Memoization Explanation?
So, your code can be improved like the folloiwng, by adapting memoization algorithm.
var n = function getRandomNum() {
return Math.floor(Math.random()*100) +1;
}
var result = [];
result[0] = 1;
result[1] = 1;
function fib(x) {
var ix, ixLen;
for(ix = 0, ixLen = x; ix < ixLen; ix++){
if(!result[ix]){
result[ix] = result[ix-2] + result[ix-1];
}
}
console.log('n:', x, ' result: ', result[ix-1]);
return result[ix-1];
}
console.log(fib(n()));
Compare the result with this site.
http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fibonacci/fibtable.html

Get factorial of all the numbers of array recursively

I am trying to get factorial of all the numbers of array(recurArray) using recursion and without loops.
I am getting Error "Maximum call stack size exceeded"
I think there is some issue with the for loop logic, would be helpful if someone can explain the cause of error and how to fix it
Thanks.
//code
function recur(){
var n;
var result;
if(n == 1)
return 1;
var recurArray = [5,6,7,8,9];
for (var i = 0;i<recurArray.length;i++){
n = recurArray[i];
result = n * recur(n-1);
n=n-1;
}
console.log("val of n " + n + "value of i " + i);
return result;
}
recur();

Your recur() function should probably take n as an argument, otherwise n will never be 1 (if(n == 1) return 1;) and your function will keep calling itself until it crashes.
Try function recur(n){ instead.

As you have array, you should use loop.
function recur(x) {
if(x==0) {
return 1;
}
return x * recur(x-1);
}
function getFact() {
var recurArray = [5,6,7,8,9];
for (var i = 0;i<recurArray.length;i++){
console.log(recur(recurArray[i]));
}
}
getFact();

In Each occurence you reset the factorial so it keep rolling for fact(5) you need to have a function that calculate the factorial and an other one for the loop over your array like this :
function recur(n){
if(n == 1){
return 1;
} else {
return n* recur(n-1);
}
}
var recurArray = [5,6,7,8,9];
for (var i = 0;i<recurArray.length;i++){
n = recurArray[i];
result = recur(n);
console.log("factorial of n " + n + " is " + result);
}

Try something like this:
function factorial(number) {
var temp;
if(number <= 1) return 1;
temp = number * factorial(number - 1);
return temp;
}
In this case factorial(5); will return !5 . Recursive functions are not supposed to have loops inside(They can, but the execution time would be horrendous).
Also recursive functions call themselves with different parameters(otherwise you would overflow the browser stack). In your case you call recursive() an infinite amount of times and the loop always starts from 5 , infinitely. The passed parameter is what stops the recursion.

JavaScript Help(It's pretty simple im a beginner)

function mathProb() {
var x = parseInt(prompt("Enter first integer", ""));
var y = parseInt(prompt("Enter the second integer", ""));
var operand = prompt("Enter type of operation", "");
if (operand == "+" || "add") {
var sum = x + y;
document.write("Your sum is " + sum);
} else if (operand == "-") {
var difference = x - y;
document.write("Your difference is " + difference);
} else if (operand == "*") {
var product = x * y;
document.write("Your product is " + product);
} else if (operand == "/") {
var quotient = x / y;
document.write("Your quotient is " + quotient);
} else {
document.write("Oops something went wrong");
}
}
Well to start I am reading a book on JavaScript and have been doing pretty well, I am now on functions and was getting those until parameters were introduced can someone explain what a parameter is in a clear simple way?
Why does this function work when named function mathProb() and function mathProb(x,y,operand)?
And a third question off of the previous one is why when I call the function in html
(<input type="button" value="Calculator" onclick="mathProb()"/>)
I have to use mathProb() even if its named mathProb(x,y,operand). If I call it using that name it wont work. Please help?

First, the line:
if(operand=="+"||"add")
Will always be true, as the expression "add" will always return a true-ish value. You probably mean to use:
if(operand=="+" || operand=="add")
Your question about parameters is probably a pretty broad topic. Basically, a parameter is a variable given to a function so that the function can be generalized to work with any data. For example, if you wanted to write a function that can add two numbers, the function must know which two numbers to add. These numbers would be supplied as parameters:
function add(x, y)
{
return x + y; // x and y are variables known within this function
}
You'd then call your function as so:
var oneplusone = add(1, 1); // Adds 1 and 1
Using this knowledge, you could rewrite your code as this:
function mathProb(x, y, operand)
{
// No need for var x, etc as these can now be passed in..
}
Then call your function:
mathProb(
parseInt(prompt("Enter first integer","")), // This is x
parseInt(prompt("Enter the second integer","")), // This is y
prompt("Enter type of operation","") // This is operand
);
Keep in mind you could still call your function mathProb without the parameters:
mathProb();
...if you really wanted to. JavaScript does allow this (unlike many other languages). However, within your function, the variables x, y and operand will be undefined which might cause unexpected results if you don't account for that.

You need call and pass function like mathProb(1,2,'+')
HTML:
<input type="button" value="Calculator" onclick="mathProb(1,2,'+')"/>
Javacript:
function mathProb(x,y,operand)
{
//var x = parseInt(prompt("Enter first integer",""));
//var y = parseInt(prompt("Enter the second integer",""));
//var operand = prompt("Enter type of operation","");
if(operand=="+"|| operand=="add")
{
var sum = x+y;
document.write("Your sum is " +sum);
}
else if(operand=="-")
{
var difference = x-y;
document.write("Your difference is " +difference);
}
else if(operand=="*")
{
var product = x*y;
document.write("Your product is " +product);
}
else if(operand=="/")
{
var quotient = x/y;
document.write("Your quotient is " +quotient);
}
else
{
document.write("Oops something went wrong");
}
}

How can I make var a = add(2)(3); //5 work?

I want to make this syntax possible:
var a = add(2)(3); //5
based on what I read at http://dmitry.baranovskiy.com/post/31797647
I've got no clue how to make it possible.

You need add to be a function that takes an argument and returns a function that takes an argument that adds the argument to add and itself.
var add = function(x) {
return function(y) { return x + y; };
}

function add(x) {
return function(y) {
return x + y;
};
}
Ah, the beauty of JavaScript
This syntax is pretty neat as well
function add(x) {
return function(y) {
if (typeof y !== 'undefined') {
x = x + y;
return arguments.callee;
} else {
return x;
}
};
}
add(1)(2)(3)(); //6
add(1)(1)(1)(1)(1)(1)(); //6

It's about JS curring and a little strict with valueOf:
function add(n){
var addNext = function(x) {
return add(n + x);
};
addNext.valueOf = function() {
return n;
};
return addNext;
}
console.log(add(1)(2)(3)==6);//true
console.log(add(1)(2)(3)(4)==10);//true
It works like a charm with an unlimited adding chain!!

function add(x){
return function(y){
return x+y
}
}
First-class functions and closures do the job.

function add(n) {
sum = n;
const proxy = new Proxy(function a () {}, {
get (obj, key) {
return () => sum;
},
apply (receiver, ...args) {
sum += args[1][0];
return proxy;
},
});
return proxy
}
Works for everything and doesn't need the final () at the end of the function like some other solutions.
console.log(add(1)(2)(3)(10)); // 16
console.log(add(10)(10)); // 20

try this will help you in two ways add(2)(3) and add(2,3)
1.)
function add(a){ return function (b){return a+b;} }
add(2)(3) // 5
2.)
function add(a,b){
var ddd = function (b){return a+b;};
if(typeof b =='undefined'){
return ddd;
}else{
return ddd(b);
}
}
add(2)(3) // 5
add(2,3) // 5

ES6 syntax makes this nice and simple:
const add = (a, b) => a + b;
console.log(add(2, 5));
// output: 7
const add2 = a => b => a + b;
console.log(add2(2)(5));
// output: 7

Arrow functions undoubtedly make it pretty simple to get the required result:
const Sum = a => b => b ? Sum( a + b ) : a;
console.log(Sum(3)(4)(2)(5)()); //14
console.log(Sum(3)(4)(1)()); //8

This is a generalized solution which will solve add(2,3)(), add(2)(3)() or any combination like add(2,1,3)(1)(1)(2,3)(4)(4,1,1)(). Please note that few security checks are not done and it can be optimized further.
function add() {
var total = 0;
function sum(){
if( arguments.length ){
var arr = Array.prototype.slice.call(arguments).sort();
total = total + arrayAdder(arr);
return sum;
}
else{
return total;
}
}
if(arguments.length) {
var arr1 = Array.prototype.slice.call(arguments).sort();
var mytotal = arrayAdder(arr1);
return sum(mytotal);
}else{
return sum();
}
function arrayAdder(arr){
var x = 0;
for (var i = 0; i < arr.length; i++) {
x = x + arr[i];
};
return x;
}
}
add(2,3)(1)(1)(1,2,3)();

This will handle both
add(2,3) // 5
or
add(2)(3) // 5
This is an ES6 curry example...
const add = (a, b) => (b || b === 0) ? a + b : (b) => a + b;

This is concept of currying in JS.
Solution for your question is:
function add(a) {
return function(b) {
return a + b;
};
}
This can be also achieved using arrow function:
let add = a => b => a + b;
solution for add(1)(2)(5)(4)........(n)(); Using Recursion
function add(a) {
return function(b){
return b ? add(a + b) : a;
}
}
Using ES6 Arrow function Syntax:
let add = a => b => b ? add(a + b) : a;

in addition to what's already said, here's a solution with generic currying (based on http://github.com/sstephenson/prototype/blob/master/src/lang/function.js#L180)
Function.prototype.curry = function() {
if (!arguments.length) return this;
var __method = this, args = [].slice.call(arguments, 0);
return function() {
return __method.apply(this, [].concat(
[].slice.call(args, 0),
[].slice.call(arguments, 0)));
}
}
add = function(x) {
return (function (x, y) { return x + y }).curry(x)
}
console.log(add(2)(3))

Concept of CLOSURES can be used in this case.
The function "add" returns another function. The function being returned can access the variable in the parent scope (in this case variable a).
function add(a){
return function(b){
console.log(a + b);
}
}
add(2)(3);
Here is a link to understand closures http://www.w3schools.com/js/js_function_closures.asp

const add = a => b => b ? add(a+b) : a;
console.log(add(1)(2)(3)());
Or (`${a} ${b}`) for strings.

With ES6 spread ... operator and .reduce function. With that variant you will get chaining syntax but last call () is required here because function is always returned:
function add(...args) {
if (!args.length) return 0;
const result = args.reduce((accumulator, value) => accumulator + value, 0);
const sum = (...innerArgs) => {
if (innerArgs.length === 0) return result;
return add(...args, ...innerArgs);
};
return sum;
}
// it's just for fiddle output
document.getElementById('output').innerHTML = `
<br><br>add() === 0: ${add() === 0 ? 'true' : 'false, res=' + add()}
<br><br>add(1)(2)() === 3: ${add(1)(2)() === 3 ? 'true' : 'false, res=' + add(1)(2)()}
<br><br>add(1,2)() === 3: ${add(1,2)() === 3 ? 'true' : 'false, res=' + add(1,2)()}
<br><br>add(1)(1,1)() === 3: ${add(1)(1,1)() === 3 ? 'true' : 'false, res=' + add(1)(1,1)()}
<br><br>add(2,3)(1)(1)(1,2,3)() === 13: ${add(2,3)(1)(1)(1,2,3)() === 13 ? 'true' : 'false, res=' + add(2,3)(1)(1)(1,2,3)()}
`;
<div id='output'></div>

can try this also:
let sum = a => b => b ? sum(a + b) :a
console.log(sum(10)(20)(1)(32)()) //63

const sum = function (...a) {
const getSum = d => {
return d.reduce((i,j)=> i+j, 0);
};
a = getSum(a);
return function (...b) {
if (b.length) {
return sum(a + getSum(b));
}
return a;
}
};
console.log(sum(1)(2)(3)(4,5)(6)(8)())

function add(a, b){
return a && b ? a+b : function(c){return a+c;}
}
console.log(add(2, 3));
console.log(add(2)(3));

This question has motivated so many answers already that my "two pennies worth" will surely not spoil things.
I was amazed by the multitude of approaches and variations that I tried to put "my favourite" features, i. e. the ones that I would like to find in such a currying function together, using some ES6 notation:
const add=(...n)=>{
const vsum=(a,c)=>a+c;
n=n.reduce(vsum,0);
const fn=(...x)=>add(n+x.reduce(vsum,0));
fn.toString=()=>n;
return fn;
}
let w=add(2,1); // = 3
console.log(w()) // 3
console.log(w); // 3
console.log(w(6)(2,3)(4)); // 18
console.log(w(5,3)); // 11
console.log(add(2)-1); // 1
console.log(add()); // 0
console.log(add(5,7,9)(w)); // 24
.as-console-wrapper {max-height:100% !important; top:0%}
Basically, nothing in this recursively programmed function is new. But it does work with all possible combinations of arguments mentioned in any of the answers above and won't need an "empty arguments list" at the end.
You can use as many arguments in as many currying levels you want and the result will be another function that can be reused for the same purpose. I used a little "trick" to also get a numeric value "at the same time": I redefined the .toString() function of the inner function fn! This method will be called by Javascript whenever the function is used without an arguments list and "some value is expected". Technically it is a "hack" as it will not return a string but a number, but it will work in a way that is in most cases the "desired" way. Give it a spin!

Simple Recursion Solution for following use cases
add(); // 0
add(1)(2)(); //3
add(1)(2)(3)(); //6
function add(v1, sum = 0) {
if (!v1) return sum;
sum += v1
return (v2) => add(v2, sum);
}

function add() {
var sum = 0;
function add() {
for (var i=0; i<arguments.length; i++) {
sum += Number(arguments[i]);
}
return add;
}
add.valueOf = function valueOf(){
return parseInt(sum);
};
return add.apply(null,arguments);
}
// ...
console.log(add() + 0); // 0
console.log(add(1) + 0);/* // 1
console.log(add(1,2) + 0); // 3

function A(a){
return function B(b){
return a+b;
}
}
I found a nice explanation for this type of method. It is known as Syntax of Closures
please refer this link
Syntax of Closures

Simply we can write a function like this
function sum(x){
return function(y){
return function(z){
return x+y+z;
}
}
}
sum(2)(3)(4)//Output->9

Don't be complicated.
var add = (a)=>(b)=> b ? add(a+b) : a;
console.log(add(2)(3)()); // Output:5
it will work in the latest javascript (ES6), this is a recursion function.

Here we use concept of closure where all the functions called inside main function iter refer and udpate x as they have closure over it. no matter how long the loop goes , till last function , have access to x.
function iter(x){
return function innfunc(y){
//if y is not undefined
if(y){
//closure over ancestor's x
x = y+x;
return innfunc;
}
else{
//closure over ancestor's x
return x;
}
}
}
iter(2)(3)(4)() //9
iter(1)(3)(4)(5)() //13

let multi = (a)=>{
return (b)=>{
return (c)=>{
return a*b*c
}
}
}
multi (2)(3)(4) //24
let multi = (a)=> (b)=> (c)=> a*b*c;
multi (2)(3)(4) //24

we can do this work using closure.
function add(param1){
return function add1(param2){
return param2 = param1 + param2;
}
}
console.log(add(2)(3));//5

I came up with nice solution with closure, inner function have access to parent function's parameter access and store in its lexical scope, when ever we execute it, will get answer
const Sum = function (a) {
return function (b) {
return b ? Sum(a + b) : a;
}
};
Sum(1)(2)(3)(4)(5)(6)(7)() // result is 28
Sum(3)(4)(5)() // result is 12
Sum(12)(10)(20) // result is 42
enter image description here

You should go in for currying to call the function in the above format.
Ideally, a function which adds two numbers will be like,
let sum = function(a, b) {
return a + b;
}
The same function can be transformed as,
let sum = function(a) {
return function(b) {
return a+b;
}
}
console.log(sum(2)(3));
Let us understand how this works.
When you invoke sum(2), it returns
function(b) {
return 2 + b;
}
when the returned function is further invoked with 3, b takes the value 3. The result 5 is returned.
More Detailed Explanation:
let sum = function(a) {
return function(b) {
return a + b;
}
}
let func1 = sum(2);
console.log(func1);
let func2 = func1(3)
console.log(func2);
//the same result can be obtained in a single line
let func3 = sum(2)(3);
console.log(func3);
//try comparing the three functions and you will get more clarity.

This is a short solution:
const add = a => b => {
if(!b) return a;
return add(a + b);
}
add(1)(2)(3)() // 6
add(1)(2)(3)(4)(5)() // 15